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Express 8 cos θ - 6 sin θ in the form of Rsin(θ-α), where R is positive and α is an acute angle in radians.

a) 10sin(θ - π/3)
b) 10sin(θ + π/3)
c) 10sin(θ - π/6)
d) 10sin(θ + π/6)

1 Answer

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Final answer:

To express 8 cos θ - 6 sin θ as Rsin(θ-α), we find R as 10 and α as π/3, so the expression becomes a) 10sin(θ - π/3).

Step-by-step explanation:

To express 8 cos θ - 6 sin θ in the form Rsin(θ-α), we need to find R and α where R is positive and α is an acute angle in radians.

First, let's find R using the Pythagorean theorem:

R = √(8^2 + 6^2) = √(64 + 36) = √100 = 10.

Now, let's determine α such that:

cos(α) = 8/R = 8/10 = 0.8

sin(α) = 6/R = 6/10 = 0.6

α = arctan(6/8) = arctan(0.75) = π/4.

However, we need a transformation that satisfies the expressions for both cosine and sine. If we look at the answer options, we know that α must be either π/3 or π/6 because these values correspond to familiar values of sin and cos.

Since cos(π/6) = 0.866 and sin(π/6) = 0.5, these do not match the required values. However, cos(π/3) = 0.5 and sin(π/3) = 0.866, which, when multiplied by R (10), give us the coefficients we started with (8 and 6, respectively, but the sin coefficient is negative).

Hence the correct expression is 10sin(θ - π/3), which corresponds to option (a).

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